Problem 49: Prime permutations

The arithmetic sequence, 1487, 4817, 8147, in which each of the terms increases by 3330, is unusual in two ways:
(i) each of the three terms are prime, and,
(ii) each of the 4-digit numbers are permutations of one another.

There are no arithmetic sequences made up of three 1-, 2-, or 3-digit primes, exhibiting this property, but there is one other 4-digit increasing sequence.

What 12-digit number do you form by concatenating the three terms in this sequence?

My Algorithm

My function fingerprint counts how often each digit occurs and produces an integer (which may have up to 10 digits).
The n-th decimal digit of the result represents how often the digit n occurs in the input, e.g.fingerprint(454430) = 131001
because 5 appears once, 4 three times, 3 once, no 2, no 1 and a single zero.fingerprint has the nice property that two number with the same fingerprint are a permutation of each other.

After generating all primes, their fingerprints are stored.

All permutations of a prime number, which are prime themselves, will be added to a list of candidates.
There must be at least sequenceLength (it's 3 in the original problem) candidates.

However, some candidates may not have the proper distance to each other: that's why I compute the differences of each candidate prime to each other.
Only if at least sequenceLength primes share the same distance to at least one other prime, then we may have a result.
Unfortunately, pairs may have the same distance diff = |p_a - p_b| = |p_c - p_d| but are disjunct: diff != |p_a - p_c|.
For example, the primes 3, 5, 17, 19 have a pair-wise distance of 2 (3-5 and 17-19) but there is no way to connect 3 and 5 to 17 and 19.
Therefore the program tries to start at every candidate prime p_i and looks for the longest sequence p_i + diff, p_i + 2 * diff, p_i + 3 * diff, ...
where each element p_i + k * diff is part of the candidates.

If such a sequence was found, the program repeats the same process but connects all elements to a long string.
All strings are stored in an std::set which is automatically ordered.

Substantial parts of my code are due to Hackerrank's modifications: the sequenceLength may be 3 or 4 and a user-defined upper limit exists.
Default values for the original problem would be 10000 and 3.

Interactive test

You can submit your own input to my program and it will be instantly processed at my server:

This live test is based on the Hackerrank problem.

Input data (separated by spaces or newlines):

This is equivalent toecho "2000 3" | ./49

Output:

(please click 'Go !')

(this interactive test is still under development, computations will be aborted after one second)

My code

… was written in C++11 and can be compiled with G++, Clang++, Visual C++. You can download it, too.

The code contains #ifdefs to switch between the original problem and the Hackerrank version.Enable #ifdef ORIGINAL to produce the result for the original problem (default setting for most problems).

#include<set>

#include<map>

#include<iostream>

#include<string>

#include<algorithm>

// count how often each digit appears: result's n-th digit describes how often n appears in x

// e.g. 454430 => 131001

// because 5 appears once, 4 three times, 3 once, no 2, no 1 and a single zero

unsignedlonglongfingerprint(unsignedint x)

{

unsignedlonglong result = 0;

while (x > 0)

{

auto digit = x % 10;

x /= 10;

unsignedlonglong pos = 1;

for (unsignedint i = 1; i <= digit; i++)

pos *= 10;

result += pos;

}

return result;

}

intmain()

{

unsignedint limit = 10000;

unsignedint sequenceLength = 4;

std::cin>> limit >> sequenceLength;

// find primes (simple sieve)

std::set<unsignedint> primes;

primes.insert(2);

for (unsignedint i = 3; i < 1000000; i += 2)

{

bool isPrime = true;

for (auto p : primes)

{

// next prime is too large to be a divisor ?

if (p*p > i)

break;

// divisible ? => not prime

if (i % p == 0)

{

isPrime = false;

break;

}

}

// yes, we have a prime number

if (isPrime)

primes.insert(i);

}

// count fingerprints of all prime numbers

std::map<unsignedlonglong, unsignedint> fingerprints;

for (auto p : primes)

fingerprints[fingerprint(p)]++;

// [length] => [merged primes, alphabetically ordered]

std::map<unsignedint, std::set<std::string>> result;

// iterate through all primes

for (auto p : primes)

{

// at least three digits ...

if (p < 1000)

continue;

// but not too far ...

if (p >= limit)

break;

// too few primes sharing this fingerprint ?

if (fingerprints[fingerprint(p)] < 3)

continue;

// generate all digit permutations

std::string digits = std::to_string(p);

std::sort(digits.begin(), digits.end());

// find all permutations which are primes

std::set<unsignedint> candidates;

do

{

// first digit can't be zero

if (digits[0] =='0')

continue;

// convert to an integer

unsignedint permuted = std::stoi(digits);

// permutation must be prime, too

if (primes.count(permuted) == 0)

continue;

// we already had this sequence ?

if (permuted < p)

break;

// yes, a valid prime

candidates.insert(permuted);

} while (std::next_permutation(digits.begin(), digits.end()));

// too few candidates ?

if (candidates.size() < sequenceLength)

continue;

// compute differences of each prime to each other prime

// [difference] => [primes that are that far away from another prime]

std::map<unsignedint, std::set<unsignedint>> differences;

for (auto bigger : candidates)

for (auto smaller : candidates)

{

// ensure smaller < bigger

if (smaller >= bigger)

break;

// store both primes

differences[bigger - smaller].insert(bigger);

differences[bigger - smaller].insert(smaller);

}

// walk through all differences

for (auto d : differences)

{

// at least 3 or 4 numbers must be involved in a sequence

if (d.second.size() < sequenceLength)

continue;

// current difference

auto diff = d.first;

// potential numbers for a sequence

auto all = d.second;

// could be a false alarm if disjunct pairs have the same difference

// we need a sequence ...

for (auto start : all)

{

// out of bounds ?

if (start >= limit)

continue;

// count numbers which can be reached by repeatedly adding our current difference

unsignedint followers = 0;

unsignedint next = start + diff;

while (all.count(next) != 0)

{

followers++;

next += diff;

}

// found enough ? => print result

if (followers >= sequenceLength - 1)

{

// same loop as before, but this time we merge the numbers into a string

Changelog

Hackerrank

Difficulty

5%
Project Euler ranks this problem at 5% (out of 100%).

Hackerrank describes this problem as hard.

Note:Hackerrank has strict execution time limits (typically 2 seconds for C++ code) and often a much wider input range than the original problem.In my opinion, Hackerrank's modified problems are usually a lot harder to solve. As a rule thumb: brute-force is rarely an option.

Those links are just an unordered selection of source code I found with a semi-automatic search script on Google/Bing/GitHub/whatever.
You will probably stumble upon better solutions when searching on your own.
Maybe not all linked resources produce the correct result and/or exceed time/memory limits.

Heatmap

Please click on a problem's number to open my solution to that problem:

green

solutions solve the original Project Euler problem and have a perfect score of 100% at Hackerrank, too

yellow

solutions score less than 100% at Hackerrank (but still solve the original problem easily)

gray

problems are already solved but I haven't published my solution yet

blue

solutions are relevant for Project Euler only: there wasn't a Hackerrank version of it (at the time I solved it) or it differed too much

orange

problems are solved but exceed the time limit of one minute or the memory limit of 256 MByte

red

problems are not solved yet but I wrote a simulation to approximate the result or verified at least the given example - usually I sketched a few ideas, too

black

problems are solved but access to the solution is blocked for a few days until the next problem is published

[new]

the flashing problem is the one I solved most recently

I stopped working on Project Euler problems around the time they released 617.

The 310 solved problems (that's level 12) had an average difficulty of 32.6&percnt; at Project Euler and
I scored 13526 points (out of 15700 possible points, top rank was 17 out of &approx;60000 in August 2017)
at Hackerrank's Project Euler+.

My username at Project Euler is stephanbrumme while it's stbrumme at Hackerrank.

Copyright

I hope you enjoy my code and learn something - or give me feedback how I can improve my solutions.All of my solutions can be used for any purpose and I am in no way liable for any damages caused.You can even remove my name and claim it's yours. But then you shall burn in hell.

The problems and most of the problems' images were created by Project Euler.Thanks for all their endless effort !!!

more about me can be found on my homepage,
especially in my coding blog.
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